📂Measure Theory

Proof of the Karatheodory's Theorem

Definition¹

For all $A \subset X$, if the following equation holds, then $E \subset X$ is said to satisfy the Caratheodory condition, or $E$ is said to be $\mu^{\ast}$-measurable.

$$\begin{equation} \mu^{\ast}(A) = \mu^{\ast}(A\cap E) + \mu^{\ast}(A \cap E^{c}) \label{def1} \end{equation}$$

$\mu^{\ast}$ is an outer measure.

Theorem

Let $L$ be the set that includes all $E \subset X$ that satisfy the Caratheodory condition. Then, $L$ is a $\sigma$-algebra. Also, $\mu^{\ast}$ becomes a measure on $L$.

This is known as the Caratheodory theorem.

Explanation

If we say $X=\mathbb{R}^n$, then $L$ is called the Lebesgue $\sigma$-algebra of $\mathbb{R}^n$. And, $E$ is called the Lebesgue measurable set or simply measurable set of $\mathbb{R}^n$. And at this time, the outer measure $\mu^{\ast}$ is called the Lebesgue measure on $\mathbb{R}^n$.

Proof

To show that $\mu^{\ast}$ becomes a measure on $L$, it suffices to check if the following three conditions are met by definition.

Measure

(D1) $\mu^{\ast}(\varnothing)=0$

(D2) $\mu^{\ast} : L \to [0,\infty]$

(D3) For distinct $E_{i} \subset X$, $\mu^{\ast}\left(\bigsqcup \limits_{i=1}^{\infty} E_{i} \right) = \sum \limits_{i=1}^{\infty}\mu^{\ast}(E_{i})$

However, this is trivially satisfied by the definition and properties of the outer measure.

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To show that $L$ is a $\sigma$-algebra, it suffices to check if the following conditions are met by definition.

$\sigma$-algebra

Let a set $X$ be given. The collection $\mathcal{E} \subset \mathcal{P}(X)$ of subsets of $X$ that satisfies the following conditions is called a $\sigma$-algebra.

• (D1) $\varnothing, X \in \mathcal{E}$
• (D2) $E \in \mathcal{E} \implies E^{c} \in \mathcal{E}$
• (D3) $E_{k} \in \mathcal{E}\ (\forall k \in \mathbb{N}) \implies \bigcup_{k=1}^\infty E_{k} \in \mathcal{E}$

• (D1)

  By substituting $E$ with $\varnothing$, and $X$, it’s easy to verify that it holds based on the Caratheodory condition.

• (D2)

  This trivially holds by the definition of the Caratheodory condition.

• (D3)

  Unlike the previous two conditions, this is not straightforward to verify. First, it will be shown that when $E_{1}$ and $E_{2}$ satisfy $\eqref{def1}$, then $E_{1} \cup E_{2}$ also satisfies $\eqref{def1}$.

  • Part 1.

    Since $E_{1}$ satisfies $\eqref{def1}$ by assumption, the following holds:

    $$\mu^{\ast}(A)=\mu^{\ast}(A \cap E_{1}) + \mu^{\ast}(A \cap E_{1}^{c})$$

    Similarly, since $E_{2}$ also satisfies $\eqref{def1}$, the following holds:

    $$\begin{align*} \mu^{\ast}(A) &= \mu^{\ast}(A \cap E_{1}) + \mu^{\ast}(A \cap E_{1}^{c}) \\ &= \mu^{\ast} \big( (A \cap E_{1})\cap E_{2} \big) + \mu^{\ast} \big( (A \cap E_{1})\cap E_{2}^{c} \big) \\ &\quad + \mu^{\ast} \big( (A \cap E_{1}^{c})\cap E_2 \big)+\mu^{\ast} \big( (A \cap E_{1}^{c})\cap E_{2}^c \big) \end{align*}$$

    By the countable subadditivity of the outer measure, the following equation holds:

    $$\mu^{\ast} \Big( \big[ A\cap E_{1}\cap E_{2} \big] \cup \big[ A \cap E_{1} \cap E_{2}^{c}\big] \cup \big[ A \cap E_{1}^{c} \cap E_2 \big] \Big) \\ \le \mu^{\ast} \big( A \cap E_{1}\cap E_2 \big) + \mu^{\ast} \big( A \cap E_{1}\cap E_{2}^{c} \big) + \mu^{\ast} \big( A \cap E_{1}^{c}\cap E_2 \big)$$

    Thus, the following holds:

    $$\begin{align*} & \mu^{\ast}(A) \\ \ge& \mu^{\ast} \Big( \big[ A\cap E_{1}\cap E_{2} \big] \cup \big[ A \cap E_{1} \cap E_{2}^{c}\big] \cup \big[ A \cap E_{1}^{c} \cap E_2 \big] \Big) + \mu^{\ast} (A \cap E_{1}^{c} \cap E_{2}^c ) \\ =&\ \mu^{\ast} \big( A\cap \big[ (E_{1}\cap E_{2}) \cup (E_{1} \cap E_{2}^{c}) \cup ( E_{1}^{c} \cap E_{2}) \big] \Big) + \mu^{\ast} (A \cap E_{1}^{c} \cap E_{2}^c ) \\ =&\ \mu^{\ast} \big( A\cap (E_{1} \cup E_{2}) \big) + \mu^{\ast} (A \cap E_{1}^{c} \cap E_{2}^c ) \\ =&\ \mu^{\ast} \big( A\cap (E_{1} \cup E_{2}) \big) + \mu^{\ast} \big(A \cap (E_{1} \cup E_2 )^{c} \big) \end{align*}$$

    Therefore, when $E_{1}$ and $E_{2}$ satisfy $\eqref{def1}$, $E_{1}\cup E_{2}$ also satisfies $\eqref{def1}$. By repeating this, it can be understood that any given $N$ distinct $E_{i}$s satisfying $\eqref{def1}$ implies that $\bigsqcup_{i=1}^{N} E_{i}$ also satisfies $\eqref{def1}$.

  • Part 2.

    The initial condition lacks the distinctness of each $E_{i}$, which is necessary for the proof, so a bit of trickery is needed. First, define each $\tilde{E}_{i}$ as follows:

    $$\begin{align*} \tilde{E}_{1} &= E_{1} \\ \tilde{E}_{2} &= E_{2} \cap \tilde{E}_{1}^{c} \\ \tilde{E}_{3} &= E_{3} \cap (\tilde{E}_{1} \cup \tilde{E}_{2} )^{c} \\ \tilde{E}_{4} &= E_{4} \cap (\tilde{E}_{1} \cup \tilde{E}_{2} \cup \tilde{E}_{3})^{c} \\ &\vdots \end{align*}$$

    Then, each of the $\tilde{E}_{i}$s are disjoint, satisfy $\eqref{def1}$, and the following holds:

    $$\bigsqcup_{i=1}^\infty \tilde{E}_{i} = \bigsqcup _{i=1}^\infty E_{i}$$

    This can be easily verified by direct calculation. Therefore, it can now be shown without loss of generality that when each of the $\tilde{E}_{i}$s satisfies $\eqref{def1}$, $\bigsqcup_{i=1}^\infty \tilde{E}_{i}$ also satisfies $\eqref{def1}$, completing the proof.

  • Part 3.

    By the countable subadditivity, the following holds:

    $$\mu^{\ast} (A) \le \mu^{\ast}\left(A \cap \left( \bigsqcup _{i=1}^\infty \tilde{E}_{i}\right) \right) + \mu^{\ast} \left(A \cap \left( \bigsqcup _{i=1}^\infty \tilde{E}_{i} \right)^{c} \right)$$

    Showing the opposite direction of inequality completes the proof. Since it was shown in Part 1. that it holds for $N$, the following holds:

    $$\begin{align*} \mu^{\ast} (A) &= \mu^{\ast} \left( A \cap \left( \bigsqcup _{i=1}^N \tilde{E}_{i}\right) \right) + \mu^{\ast} \left(A \cap \left( \bigsqcup _{i=1}^N \tilde{E}_{i} \right)^{c} \right) \\ &\ge \mu^{\ast} \left( A \cap \left( \bigsqcup _{i=1}^N \tilde{E}_{i}\right) \right) + \mu^{\ast} \left(A \cap \left( \bigsqcup _{i=1}^\infty \tilde{E}_{i} \right)^{c} \right) \\ &= \sum \limits_{i=1}^N\mu^{\ast}\left(A \cap \tilde{E}_{i} \right) + \mu^{\ast} \left(A \cap \left( \bigsqcup _{i=1}^\infty \tilde{E}_{i} \right)^{c} \right) \end{align*}$$

    This inequality holds for all $N$, so the following equation holds:

    $$\begin{align*} \mu^{\ast} (A) & \ge \sum \limits_{i=1}^\infty \mu^{\ast}\left(A \cap \tilde{E}_{i} \right) + \mu^{\ast} \left(A \cap \left( \bigsqcup _{i=1}^\infty \tilde{E}_{i} \right)^{c} \right) \\ &\ge \mu^{\ast} \left( A \cap \left( \bigsqcup_{i=1}^\infty \tilde{E}_{i} \right) \right) + \mu^{\ast} \left(A \cap \left( \bigsqcup _{i=1}^\infty \tilde{E}_{i} \right)^{c} \right) \end{align*}$$

    The second inequality holds by countable subadditivity.

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